Integral of 9*x*e^(3*x) dx

1. The integral of a constant times a function is the constant times the integral of the function:

   $$\int 9 x e^{3 x}\, dx = 9 \int x e^{3 x}\, dx$$

   1. Use integration by parts:

      $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

      Let $u{\left(x \right)} = x$ and let $\operatorname{dv}{\left(x \right)} = e^{3 x}$.

      Then $\operatorname{du}{\left(x \right)} = 1$.

      To find $v{\left(x \right)}$:

      1. There are multiple ways to do this integral.

         Method #1

         1. Let $u = 3 x$.

            Then let $du = 3 dx$ and substitute $\frac{du}{3}$:

            $$\int \frac{e^{u}}{9}\, du$$

            1. The integral of a constant times a function is the constant times the integral of the function:

               $$\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}$$

               1. The integral of the exponential function is itself.

                  $$\int e^{u}\, du = e^{u}$$

               So, the result is: $\frac{e^{u}}{3}$

            Now substitute $u$ back in:

            $$\frac{e^{3 x}}{3}$$

         Method #2

         1. Let $u = e^{3 x}$.

            Then let $du = 3 e^{3 x} dx$ and substitute $\frac{du}{3}$:

            $$\int \frac{1}{9}\, du$$

            1. The integral of a constant times a function is the constant times the integral of the function:

               $$\int \frac{1}{3}\, du = \frac{\int 1\, du}{3}$$

               1. The integral of a constant is the constant times the variable of integration:

                  $$\int 1\, du = u$$

               So, the result is: $\frac{u}{3}$

            Now substitute $u$ back in:

            $$\frac{e^{3 x}}{3}$$

      Now evaluate the sub-integral.

   2. The integral of a constant times a function is the constant times the integral of the function:

      $$\int \frac{e^{3 x}}{3}\, dx = \frac{\int e^{3 x}\, dx}{3}$$

      1. Let $u = 3 x$.

         Then let $du = 3 dx$ and substitute $\frac{du}{3}$:

         $$\int \frac{e^{u}}{9}\, du$$

         1. The integral of a constant times a function is the constant times the integral of the function:

            $$\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}$$

            1. The integral of the exponential function is itself.

               $$\int e^{u}\, du = e^{u}$$

            So, the result is: $\frac{e^{u}}{3}$

         Now substitute $u$ back in:

         $$\frac{e^{3 x}}{3}$$

      So, the result is: $\frac{e^{3 x}}{9}$

   So, the result is: $3 x e^{3 x} - e^{3 x}$

2. Now simplify:

   $$\left(3 x - 1\right) e^{3 x}$$

3. Add the constant of integration:

   $$\left(3 x - 1\right) e^{3 x}+ \mathrm{constant}$$

The answer is:

$$\left(3 x - 1\right) e^{3 x}+ \mathrm{constant}$$